Probability question (1 Viewer)

apollo1

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A family consists of a father, mother, 3 girls and 4 boys.

If same family is seated randomly at a round table, find the probability that the parents are separated exactly by two seats, and these seats are occupied by boys.

i got 1/14 can sumone confirm if its right or wrong. :)
 

michaeljennings

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A family consists of a father, mother, 3 girls and 4 boys.

If same family is seated randomly at a round table, find the probability that the parents are separated exactly by two seats, and these seats are occupied by boys.

i got 1/14 can sumone confirm if its right or wrong. :)
i got the same answer as you

tell me your method

mine was 4C2 for the two boys, times by two because they can swap positions, times by two again cos mother and father can swap positions and then 5! for the remaining children all divided by total possibilities (8!)
 
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someth1ng

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Total Arrangements=(9-1)!=40320 Arrangements
Arrangements with Conditions=(6-1)!x2x(4C2)x2=2880 Arrangements [Note the (6-1)! since it is circular arrangement taking the group of 4 as one block]
P(Conditions Met)=2880/40320=1/14
 
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apollo1

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i got the same answer as you

tell me your method

mine was 4C2 for the two boys, times by two because they can swap positions, times by two again cos mother and father can swap positions and then 5! for the remaining children all divided by total possibilities (8!)
this is wat i did as well.
 

descartes

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Another way of thinking about it;

1. Place the father in the circle, as it is a circle there is no defined positions so there is only one way way of doing this, and this is to put him in the circle. (1 way)

2. Now place the mother, as someone is sitting in the circle positions can now be defined, so ther are two ways of doing this, three seats to the left or three seats to the right. (2 ways)

Alternative for 1 & 2 combined.
Father and mother can be arranged 2! ways, place them in a circle, as it is a circle there is no defined positions so there is only one way way of doing this, and this is to put them in the circle. (2! X 1 = 2 ways)

3. Two boys must sit between, but order is important, so of the four select two boys (4P2 ways)

4. This leaves 5 children to seat (5! ways)

Total ways = 1 X 2 X 4P2 X 5!

Probability = (1 X 2X 4P2 X 5!)/8! = 1/14
 

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